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CHAPTER 6 Algebra of Complex Numbers
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It is important, now, to note that for all values of k less than n the quantity k=n is not equal to an integer, but is, instead, equal to a fraction of value less than 1. Thus, for all values of k less than n it is not true that     k 8 sin sin 360 n n n and thus we have that A complex number, a jb, has n roots, given by eq. (176), for all values of k less than n, that is, for k 0; 1; 2; 3; . . . ; n 1 .* Thus, in this section we ve found that a jb raised to a positive or negative INTEGRAL power has just one value, given by eq. (168). On the other hand, the n th root of a jb that is, a jb raised to the 1=n power consists of n distinct values, given by eq. (176). It should be noted that inverse operations tend to produce multiple answers, in the manner of eq. (176). For example, while 42 gives the single answer 16, note that p 4 2 and 2 or note that tan 458 1:000 but arctan 1:000 458; 2258; 4058; 5858; and so on: In regard to eq. (176), it might be thought that such an equation, while interesting theoretically, would have little practical value. Actually, however, the equation has numerous engineering applications; it is, for example, central in the study of broad-band cascaded ampli er stages. Problem 105 Find the four fourth roots of 3 j7 . Show, on the complex plane, the points representing the four roots. From our work in the foregoing problem, the procedure for nding the values (roots) of the complex expression a jb 1=n can be summarized as follows. The GIVEN VALUES will be the values of a, b, and n. The rst step, then, is to calculate the magnitude A and angle  of the complex number a jb , which is done by means of the formulas p 177 A a2 b2 a2 b2 1=2  arctan b=a 178
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In eq. (177), the magnitude A is always taken to be a positive, real number. In nding the value of  in eq. (178), we must take into account the quadrant that the point a, b lies in (see Fig. 110).
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* The same values are merely repeated over and over for k greater than n 1 .
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CHAPTER 6 Algebra of Complex Numbers
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As was pointed out in problem 105, the roots all have the same magnitude, A1=n , which, by eq. (177), is equal to A1=n a2 b2 1=2n 179
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Since the value of  is known (by eq. (178)), and since n is a given value, we can now nd the value of =n, which must now be substituted into eq. (176). We also know the n di erent values of k, k 0; 1; 2; . . . ; n 1 , which we must now substitute, in succession, into eq. (176). Doing this gives us the n roots of a jb 1=n . Problem 106 Find the fth roots of the complex number 19 j33 . For convenience, round o all numbers to two decimal places. Answers in polar and rectangular form. Problem 107 Find the cube roots of unity ; that is, nd the values of 1 1=3 .
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Quantities that obey the PARALLELOGRAM LAW of addition and subtraction are said to be VECTOR quantities (see note 4 in the Appendix). Hence, by this de nition, COMPLEX NUMBERS can be regarded as vector quantities, because complex numbers are basically added and subtracted in accordance with the parallelogram law. This can be shown as follows.
Fig. 112
Fig. 113
Let A and B be two complex numbers, such as are illustrated geometrically in Figs. 112 and 113. Let R be the SUM of the above two complex numbers; algebraically, from section 6.2, R is equal to R a c j b d 180
Careful inspection of the above will show that R, the SUM of the two complex numbers A and B (given by eq. (180)), can be found geometrically by drawing B o the end of A, as is done in Fig. 114.
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